Assume that a procedure yields a binomial distribution with

Assume that a procedure yields a binomial distribution with n trials and the probability of success for one trial is p. Use the given values of n and p to find the mean mu and standard deviation sigma.Also, use the range rule of thumb to find the minimum usual value mu minus 2 sigma2 and the maximum usual value mu plus 2 sigma+2.

nequals=14201420,

pequals=2 divided by 5

Solution

if X~Bin(n,p)

mean=E(X)=np

variance=var(X)=np(1-p)

so min=2=np - 2*np(1-p)

max=+2=np + 2*np(1-p)

hence for n=1420, p=2/5=.4 we get min max valus as

min=-113.6

max=1249.6